A student measures the diameter of a steel ball using a screw gauge. The true diameter of the ball is known to be 4.25 mm . The screw gauge has a pitch of 0.5 mm and 50 divisions on its circular scale. The instrument has a positive zero error of +0.04 mm . If the main scale reading during the measurement is 4.0 mm , which division of the circular scale coincides with the reference line?

Step-by-step solution

The least count (LC) of the screw gauge is given by: LC = Pitch Total circular divisions = 0.5 50 = 0.01 mm The true value of the measurement is related to the measured value and zero error by: True Value = Measured Value - Zero Error Substituting the given values: 4.25 = Measured Value - (+0.04) Measured Value = 4.25 + 0.04 = 4.29 mm The measured value is also given by: Measured Value = MSR + ( CSR LC ) Where MSR is the main scale reading and CSR is the circular scale reading. 4.29 = 4.0 + ( CSR 0.01) 0.29 = CSR 0